Problem: $\begin{aligned} f(x)&=\cos^3(x) \\\\ f'(x)&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $3\cos^2(x)$ (Choice B) B $-3\sin(x)\cos^2(x)$ (Choice C) C $-\sin^3(x)$ (Choice D) D $[1-\sin^2(x)]\cos(x)$
Answer: The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to write our function as a composite function like $w\big(u(x)\big)$. First, we realize that $\cos^3(x)$ is just shorthand for $(\cos(x))^3$, which is clearly composite: $f(x)=\underbrace{(~\overbrace{\cos(x)}^{\text{inner}}~)^3}_{\text{outer}}$ So if $f(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={\cos(x)} &&\text{inner function} \\\\ w(x)&=x^3&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-\sin(x)} \\\\ {w'(x)}&={3x^2} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={3({\cos(x)})^2} \cdot {-\sin(x)} \\\\ &=-3\sin(x)\cos^2(x) \end{aligned}$